Answer to Question #25252 in Abstract Algebra for Mohammad

Question #25252
Show that from: rad (R[t]) = (Nil*R)[t] follows Kothe’s Conjecture. (“The sum of two nil left ideals in any ring is nil”.)
Expert's answer
We shall prove Kothe’s Conjecture inthe alternative form: If R is a ring such that Nil*R = 0, thenevery nil left ideal I ⊆ R is zero. Itsuffices to show that, for any f(t) ∈ I[t], 1 + f(t) has a leftinverse in R[t], for then I[t] ⊆ rad R[t], and rad R[t] = 0 by assumption.Consider the ring R' = I ⊕ Z, obtained by formally adjoining anidentity 1' = (0, 1) to the “rng” I. It is easy to seethat Nil*R' = I, so by assumption, rad R'[t] = I[t].Given f(t) ∈ I[t],1' + f(t) has then a left inverse, say 1' + g(t),in R'[t]. From1' = (1' + g(t))(1' + f(t)),we have
−g(t) = f(t) + g(t)f(t)∈ I[t],
since I[t] is an idealin R'[t]. Now the equation 1 = (1+g(t))(1 + f(t))implies that 1 + f(t) has a left inverse 1 + g(t) ∈ R[t], as desired.

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